Expressing $\sum _{i=1}^n x_i^3\,+\,\sum _{i=1}^n\sum _{j=1}^n x_i^2 x_j$ as a polynomial in elementary symmetric polynomials I'm trying to express $\sum _{i=1}^n x_i^3\,+\,\sum _{i=1}^n\sum _{j=1}^n x_i^2 x_j$ as a polynomial in elementary symmetric polynomials.
Here is my work so far:
$$\begin{align}
(\sum _{i=1}^n x_i)^3 
&= \sum _{i,j,k=1}^n x_i x_j x_k\,+\,3\sum _{i=j\neq k}^n x_i x_j x_k\,+\,6\sum _{i\neq j\neq k,\,\,i\neq k}^n x_i x_j x_k\\ 
&=P(x)+3P_1P_2+6P_3
\end{align}$$
So $P(x)=\sum _{i=1}^n x_i^3=P_1^3-3P_1P_2-6P_3$ is a writing of $P$ in elementary symmetric polynomials.
 A: Firstly, since we are in degree three, and the degree three symmetric functions are spanned by $P_1^3$, $P_2 P_1$, and $P_3$, we should be able to express the given function as an integer-linear combination of these. A useful intermediate is the monomial symmetric functions of degree 3, which are
$$ \begin{aligned}
m_{(3)} &= \sum_i x_i^3 \\
m_{(2, 1)} &= \sum_{i \neq j} x_i^2 x_j \\
m_{(1, 1, 1)} &= P_3
\end{aligned}$$
What you have so far is that
$$ P_1^3 = m_{(3)} + 3 m_{(2, 1)} + 6 m_{(1, 1, 1)}$$
And what you are aiming at is expressed in terms of the monomial symmetric functions as 
$$ \sum_i x_i^3 + \sum_{i, j} x_i^2 x_j = 2m_{(3)} + m_{(2, 1)}$$
The rest of the problem is a matter of expanding $P_2 P_1$ in terms of the monomial basis, and doing a standard linear algebra type thing to express everything in terms of the three functions $P_1^3$, $P_2 P_1$, and $P_3$.
A: Firstly $$\sum_{i=1}^nx_i^3=\left(\sum_{i=1}^nx_i\right)^3-3\sum_{i=1}^nx_i\sum_{1\leq i<j\leq n}x_ix_j+3\sum_{1\leq i<j<k\leq n}x_ix_jx_k$$ and
$$\sum_{1\leq i<j\leq n}x_i^2x_j=\sum_{i=1}^nx_i\sum_{1\leq i<j\leq n}x_ix_j-3\sum_{1\leq i<j<k\leq n}x_ix_jx_k.$$
Id est,
$$\sum_{i=1}^nx_i^3+\sum_{i=1}^n\sum_{j=1}^nx_i^2x_j=2\sum_{i=1}^nx_i^3+\sum_{1\leq i<j\leq n}x_i^2x_j=$$
$$=2\left(\sum_{i=1}^nx_i\right)^3-5\sum_{i=1}^nx_i\sum_{1\leq i<j\leq n}x_ix_j+3\sum_{1\leq i<j<k\leq n}x_ix_jx_k.$$
Done!
